What are the solutions to the equation 0=-x^2+x+12 ?

The solutions to the equation 0=-x^2+x+12 are x=-3,x=4

Find the zeros of 0=-x^2+x+12

The zeros of 0=-x^2+x+12 are x=-3,x=4

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0 = - x^{2} + x + 12

x = - 3, x = 4

Solution steps

0 = - x^{2} + x + 12

Switch sides

- x^{2} + x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 12}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 12 = 7

x_{1, 2} = \frac{- 1 \pm 7}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 7}{2 ( - 1 )}, x_{2} = \frac{- 1 - 7}{2 ( - 1 )}

x = \frac{- 1 + 7}{2 ( - 1 )} : - 3

x = \frac{- 1 - 7}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

x = - 3, x = 4

- 5 \cdot t^{2} + 80 \cdot t = 0

- 12 x^{2} = - 32 x^{2}

12 x + 12 = 16 x^{2} - 16 x + 4

12 x^{2} = 16

(x - 5) (2 x + 3) = - 2 x - 3

What are the solutions to the equation 0=-x^2+x+12 ?
The solutions to the equation 0=-x^2+x+12 are x=-3,x=4
Find the zeros of 0=-x^2+x+12
The zeros of 0=-x^2+x+12 are x=-3,x=4